In order to facilitate the disclosure to the present subject matter, a brief discussion of thermodynamics of jet engines, magnetism and particle dynamics follow.
Fluid propulsion devices achieve thrust by imparting momentum to a fluid called the propellant. An air-breathing engine, as the name implies, uses the atmosphere for most of its propellant. The gas turbine produces high-temperature gas which may be used either to generate power for a propeller, generator or other mechanical apparatus or to develop thrust directly by expansion and acceleration of the hot gas in a nozzle. In any case, an air breathing engine continuously draws air from the atmosphere, compresses it, adds energy in the form of heat, and then expands it in order to convert the added energy to shaft work or jet kinetic energy. Thus, in addition to acting as propellant, the air acts as the working fluid in a thermodynamic process in which a fraction of the energy is made available for propulsive purposes.
The main sources of energy for air-breathing engines are hydrocarbon fuels; however, other forms of heat energy can equally be applied limited by practicality.
The performance of jet engines may be understood by means of the laws of thermodynamics, however these laws also restrict the performance to certain upper limits which depend strongly on the maximum temperature the engine can withstand.
Fairly general equations for thrust and efficiency of air breathing jet engines can be derived from the momentum and energy laws without the need for detailed consideration of the internal mechanisms of particular engines. Consider, for example, the generalized thrust producing device 100 illustrated in FIG. 1, as observed from a stationary position with respect to the device. In FIG. 1, a control surface 101 is specified which passes through the propellant outlet plane at station 2 and extends far upstream at station 1. The side surfaces of the control volume are parallel to the upstream velocity u and far removed from the thrust device 100. It is assumed for this discussion that the thrust and conditions at all points within the control volume do not vary temporally.
The reaction to the Thrust T transmitted through the structural support 102 is indicated in FIG. 1. In this sense the engine thrust may be defined as the vector summation of all forces on the internal and external surfaces of the engine and nacelle.
The thrust of the generalized thrust producer as it applies to steady flow in the x direction only gives:
                              ∑                      F            x                          =                  ∮                                                    u                x                            ⁡                              (                                  ρ                  ⁢                                                                          ⁢                                      u                    ·                    n                                                  )                                      ⁢                                          ⅆ                A                            .                                                          (        1        )            
With the assumption of reversible external flow, both the pressure and the velocity may be assumed constant over the entire control surface, except over the exhaust area Ae of the engine. If the exhaust velocity ue is supersonic, the exhaust pressure pe may differ from the ambient pressure pa. The net pressure force on the control surface is therefore (pa−pe)Ae. The only other forces acting on this control volume is the reaction to the thrust T. Adding up the forces on the control surface which act in the x-direction results in:ΣFs=(pa−pe)Ae+T.  (2)
Far upstream at station 1 the air which is drawn into the engine crosses the control surface through capture area Ai at a rate {dot over (m)}a given by {dot over (m)}a=ρuAi, in which ρ is the ambient density and u is the flight velocity. The mass flux crossing the exhaust area Ae is me=ρeueAe. Taking account of the fuel flow rate {dot over (m)}f, we have {dot over (m)}e={dot over (m)}a+{dot over (m)}f and,{dot over (m)}f=ρeueAe−ρuAi.  (3)
Now, considering the requirement of continuity for the control volume as a whole, and assuming that the fuel flow originates from outside, the control volume for steady flow is
                              ∮                      ρ            ⁢                                                  ⁢                          u              ·              n                        ⁢                          ⅆ              A                                      =        0.                            (        4        )            which for the present case may be written as:ρeueAe+ρu(A−Ai)+{dot over (m)}s−{dot over (m)}f−ρuA=0  (5)in which A is the cross sectional area of the control volume normal to the velocity u and {dot over (m)}s is the mass flow of air through the side surfaces of the control volume. However in a power generation application such as a gas turbine the mass flow of a v may be written as zero.{dot over (m)}s=ρu(Ae−Ai).  (6)
If the sides of the control volume are sufficiently distant from the thrust producer 100, it may be assumed that this flow crosses the control surfaces with a very small velocity in the γ direction and an essentially undisturbed velocity component in the x direction. Thus, the momentum carried out by the control volume with this flow is simply mu and when the components only in the x direction are considered, the right hand side of the equation may be written as:
                              ∮                                    u              x                        ⁢                          ρ              ⁡                              (                                  u                  ·                  n                                )                                      ⁢                          ⅆ              A                                      =                                                            m                .                            e                        ⁢                          u              e                                +                                                    m                .                            s                        ⁢            u                    +                      ρ            ⁢                                                  ⁢                          u              ⁡                              (                                  A                  -                                      A                    e                                                  )                                      ⁢            u                    -                                                    m                .                            a                        ⁢            u                    -                      ρ            ⁢                                                  ⁢                          u              ⁡                              (                                  A                  -                                      A                    i                                                  )                                      ⁢            u                                              (        7        )            which is the net outward flux of x-momentum from the control volume 101. Using Equation 6, we may reduce this to:
                              ∮                                    u              x                        ⁢                          ρ              ⁡                              (                                  u                  ·                  n                                )                                      ⁢                          ⅆ              A                                      =                                                            m                .                            e                        ⁢                          u              e                                -                                                    m                .                            a                        ⁢                          u              .                                                          (        8        )            
When we use Equations 2 and 8, the momentum equation 1 becomesT=T={dot over (m)}eue−{dot over (m)}sue+(pe+pa)Ae  (9)or, defining the fuel-air ration f={dot over (m)}f/{dot over (m)}a, we have:T={dot over (m)}e[(1+f)ue−u]+(pe−pa)Ae.  (10)
The term (pe−pa)Ae is non zero only if the exhaust jet is supersonic and the nozzle does not expand the exhaust jet to ambient pressure. Even if it is non zero, it is usually small compared to the momentum-flux term.
It should be borne in mind that in the derivation of Equation 10, the flow external to the engine has been assumed reversible. If this is not so, due to significant boundary layer effects such as separation, the actual force transmitted by the structural support 102 of FIG. 1 could be appreciably less than Equation 10 would predict. For engines which have two distinct exhaust streams, Equation 1 must be applied separately to each stream.